Advertisements
Advertisements
Question
\[\lim_{x \to \infty} \sqrt{x^2 + cx - x}\]
Advertisements
Solution
\[\lim_{x \to \infty} \left( \sqrt{x^2 + cx - x} \right)\]
It is of the form ∞ - ∞.
Rationalising the numerator:
\[\lim_{x \to \infty} \left( \sqrt{x^2 + cx} - x \right) \frac{\left( \sqrt{x^2 + cx} + x \right)}{\left( \sqrt{x^2 + cx} + x \right)}\]
\[ = \lim_{x \to \infty} \left( \frac{x^2 + cx - x^2}{\sqrt{x^2 + cx} + x} \right)\]
\[\text{ Dividing the numerator and the denomiator by }x: \]
\[ \lim_{x \to \infty} \left[ \frac{\frac{cx}{x}}{\frac{\sqrt{x^2 + cx} + x}{x}} \right]\]
\[ = \lim_{x \to \infty} \left[ \frac{c}{\sqrt{1 + \frac{c}{x}} + 1} \right]\]
\[\text{ When } x \to \infty , \text{ then } \frac{1}{x} \to 0 . \]
\[ \Rightarrow \frac{c}{\sqrt{1} + 1}\]
\[ = \frac{c}{2}\]
APPEARS IN
RELATED QUESTIONS
\[\lim_{x \to 3} \frac{\sqrt{2x + 3}}{x + 3}\]
\[\lim_{x \to 2} \frac{x^4 - 16}{x - 2}\]
\[\lim_{x \to 2} \left( \frac{1}{x - 2} - \frac{2}{x^2 - 2x} \right)\]
\[\lim_{x \to 1/4} \frac{4x - 1}{2\sqrt{x} - 1}\]
\[\lim_{x \to 0} \frac{\left( a + x \right)^2 - a^2}{x}\]
\[\lim_{x \to 3} \left( \frac{1}{x - 3} - \frac{3}{x^2 - 3x} \right)\]
If \[\lim_{x \to a} \frac{x^9 - a^9}{x - a} = 9,\] find all possible values of a.
\[\lim_{x \to \infty} \sqrt{x + 1} - \sqrt{x}\]
\[\lim_{x \to \infty} \frac{x}{\sqrt{4 x^2 + 1} - 1}\]
\[\lim_{x \to \infty} \left[ \left\{ \sqrt{x + 1} - \sqrt{x} \right\} \sqrt{x + 2} \right]\]
\[\lim_{x \to - \infty} \left( \sqrt{x^2 - 8x} + x \right)\]
Evaluate: \[\lim_{n \to \infty} \frac{1^4 + 2^4 + 3^4 + . . . + n^4}{n^5} - \lim_{n \to \infty} \frac{1^3 + 2^3 + . . . + n^3}{n^5}\]
\[\lim_{x \to 0} \frac{2x - \sin x}{\tan x + x}\]
\[\lim_{x \to 0} \frac{x^2 - \tan 2x}{\tan x}\]
\[\lim_{x \to 0} \frac{x^3 \cot x}{1 - \cos x}\]
\[\lim_{x \to 0} \frac{x \tan x}{1 - \cos 2x}\]
\[\lim_{x \to 0} \frac{\cos 2x - 1}{\cos x - 1}\]
\[\lim_{x \to 0} \frac{5x + 4 \sin 3x}{4 \sin 2x + 7x}\]
Evaluate the following limit:
\[\lim_{h \to 0} \frac{\left( a + h \right)^2 \sin\left( a + h \right) - a^2 \sin a}{h}\]
\[\lim_{x \to a} \frac{\cos x - \cos a}{\sqrt{x} - \sqrt{a}}\]
\[\lim_{x \to 1} \frac{1 - x^2}{\sin \pi x}\]
\[\lim_{x \to - 1} \frac{x^2 - x - 2}{\left( x^2 + x \right) + \sin \left( x + 1 \right)}\]
\[\lim_{x \to \pi} \frac{\sqrt{2 + \cos x} - 1}{\left( \pi - x \right)^2}\]
\[\lim_{x \to \pi} \frac{1 + \cos x}{\tan^2 x}\]
\[\lim_{x \to \infty} \frac{\sin x}{x} .\]
\[\lim_{x \to 0} \frac{\sin 2x}{x}\]
\[\lim_{x \to 0} \frac{\sin x^0}{x}\]
If \[\lim_{x \to 1} \frac{x + x^2 + x^3 + . . . + x^n - n}{x - 1} = 5050\] then n equal
The value of \[\lim_{x \to \infty} \frac{\sqrt{1 + x^4} + \left( 1 + x^2 \right)}{x^2}\] is
\[\lim_{x \to \pi/4} \frac{4\sqrt{2} - \left( \cos x + \sin x \right)^5}{1 - \sin 2x}\] is equal to
\[\lim_{x \to 2} \frac{\sqrt{1 + \sqrt{2 + x} - \sqrt{3}}}{x - 2}\] is equal to
The value of \[\lim_{n \to \infty} \frac{\left( n + 2 \right)! + \left( n + 1 \right)!}{\left( n + 2 \right)! - \left( n + 1 \right)!}\] is
Evaluate the following limit:
`lim_(x -> 3) [sqrt(x + 6)/x]`
Evaluate the following limits: `lim_(x -> 2)[(x^(-3) - 2^(-3))/(x - 2)]`
Number of values of x where the function
f(x) = `{{:((tanxlog(x - 2))/(x^2 - 4x + 3); x∈(2, 4) - {3, π}),(1/6tanx; x = 3"," π):}`
is discontinuous, is ______.
Evaluate the following limit:
`lim_(x->3)[sqrt(x+6)/x]`
Evaluate the following limit:
`lim_(x->7)[((root(3)(x) - root(3)(7))(root(3)(x) + root(3)(7)))/(x - 7)]`
Evaluate the following limit:
`lim _ (x -> 5) [(x^3 - 125) / (x^5 - 3125)]`
